Finding Length and Width with unknown area and perimeter

Question

I am having a bit of trouble deriving the equations for this rectangle word problem.  Here is the problem:

 

The length of a rectangle is 5 inches more than its width. The area of the rectangle is equal to 2 inches less than 2 times the perimeter. Find the length and width of the rectangle.

 

 

Answer 1

length of a rectangle is 5 inches more than its width:

L = w + 5

 

area of the rectangle is equal to 2 inches less than 2 times the perimeter:

A = 2P -2

Area = Lw

Perimeter = 2(L+w)

 

Lw = 2[2(L+w)] - 2

Lw = 4(L+w) - 2

 

Since we know L = w+5 substitute that in:

 

w(w+5) = 4(w+5+w) - 2

w² + 5w = 4(2w+5)-2

w² + 5w = 8w + 20 - 2

w² + 5w = 8w + 18

Let's move everything to left side of equal sign to create a quadratic

 

w² - 3w - 18 = 0

 

Look for factors of -18 that add to -3:

(-6)(3) = -18   and   -6+3 = -3

So this factors to:

 

(w-6)(w+3) = 0

w = 6 or w = -3

 

disregarding the negative, the width is 6 inches

L= w+5 = 11 inches:  Length is 11 inches

 

Check:  A = 2P-2

area: Lw = 66 in²

Perimeter: 2(L+w) = 2(17) = 34

66 = 2(34) - 2

66 = 68-2

66 = 66

Answer is verified